By Ivan G. Avramidi

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**Example text**

79) is satisfied. Let us also note that the form of the operator can be simplified by a similarity transformation. Let µ = e−2ω . 84) CHAPTER 2. 85) i,j=1 where n αij (∂i ω)(∂j ω) − ∂i (αij ∂j ω) . 73) can be written as follows. Suppose that the matrix αij is non-degenerate. 87) and the entries of the inverse matrix by (gij ) = (αij )−1 . 88) Let g denote the determinant of the matrix gij , g = det gij = ( det αij )−1 . 90) n g ij Ai Aj + g −1/2 ∂i (g 1/2 g ij Aj ) . 91) i,j=1 Then the operator L takes the form n g −1/2 (∂i + Ai )g 1/2 g ij (∂j + Aj ) + Q .

218) Let F (s) and G(s) be Laplace transforms of the functions f and g. 219) where σ(s) = αs2 + βs + γ . 3. 222) c−i∞ where c is a sufficiently large positive real constant. 224) c−i∞ and c+i∞ h(t) = ds st α(as + b) + aβ e . 218). These integrals can be easily computed by the residue theory. Let s1 and s2 be the poles of the integrand, in other words, the roots of the symbol σ(s), that is, the solutions of the equation σ(s) = 0 . 230) CHAPTER 2. METHODS FOR SOLUTION OF PDE 50 and 1 β 2 − 4αγ . 231) 2α Let us assume for simplicity that ∆ = 0 so that the roots are distinct.

95) Rij = ∂i Aj − ∂j Ai = 0 . 96) Thus the quantity measures the extent to which the operator L is non-self-adjoint. 3 Resolvent and Spectrum Let A be an operator on a Hilbert space. A complex number λ is called an eigenvalue of the operator A if there is a non-zero vector ϕ such that Aϕ = λϕ . 97) The vector ϕ is called the eigenvector corresponding to the eigenvalue λ. Note that there are infinitely many eigenvectors corresponding to an eigenvalue. One can easily show that the collection of all eigenvectors corresponding to a given eigenvalue λ is a vector subspace called the eigenspace) of λ.